Question 1083871: The expression 6y^2-y-51 can be rewritten as (3Ay+B)(y-C), where A, B, and C are positive integers. Find $
(AC)^2-B. Found 3 solutions by Boreal, ikleyn, MathTherapy:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! 6y^2-y-51 turns into y^2-y-306
Factor that into (y-18)(y+17), then divide the constant by 6 and reduce fully to (y-3)(y+17/6)
the factors then are (y-3) and (6y+17)
A=2
B=17
C=3
(AC)^2=36
36-17=19 is the answer.
(3Ay + B)*(y - C) = = .
Since it is identical to , we have
3A = 6 and hence A = 2; (1)
B - 3AC = -1, or B - 6C = -1; (2)
BC = 51. (3)
Thus you actually have these two equations to determine B and C:
B - 6C = -1 (2)
BC = 51. (3)
From (2), express B = 6C -1 and substitute it into (3). You will get
(6C-1)*C = 51.
= 0,
Factor left side
(C-3)*(2C+17) = 0.
Since you need C to be positive number (as the condition requires), you have only one possibility: C = 3.
Then B = 6C-1 = 6*3-1 = 17,
and now you have everything to calculate = = = 19.
You can put this solution on YOUR website!
The expression 6y^2-y-51 can be rewritten as (3Ay+B)(y-C), where A, B, and C are positive integers. Find $
(AC)^2-B.
FACTORS to: (6y + 17)(y - 3).
Since can be rewritten as: (3Ay + B)(y - C), we can then say that: (6y + 17)(y - 3) = (3Ay + B)(y - C)
By equating terms, we see that: 6y = 3Ay_____(6)y = (3A)y_____6 = 3A____
Also, B = 17, and - 3 = - C______3 = C
Thus,
